# Given $\limsup \frac{a_n}{b_n} < \infty$. Prove there is a constant M such that $a_n \leq Mb_n$ [duplicate]

Given sequences $$\{a_n\}$$ , $$\{b_n\}$$ of positive real numbers and $$\textrm{ limsup } \frac{a_n}{b_n} < \infty$$. Prove there is a constant M such that $$a_n \leq Mb_n$$

Defn: Let $$(a_{n})_{n=1}^{\infty}$$ be a bounded sequence. Define the sequence $$c_n = \sup \{a_{k}: k \geq n\}$$ for $$n \geq 1$$. If the sequence $$c_n$$ converges, then the value it converges to is the limit superior of $$(a_n)$$.

Attempt:

So given that the lim sup is finite, it means there exists a value M such that for all $$n > k$$, $$\frac{a_n}{b_n} < M \\ \Rightarrow \ a_n \leq Mb_n$$

Comment: Surely there is more to it than this and I have missed something.....

## marked as duplicate by Jeff, YuiTo Cheng, воитель, Thomas Shelby, LeucippusJun 20 at 5:30

• You haven't been explicit about where the $M$ comes from. Is it something like $\limsup + \epsilon$? And that only works for $n>k$. Your original problem was stated for all $n\in\Bbb N$. – Ted Shifrin Jun 19 at 23:43
• Still amazed that these legends of Mathematics contribute to my simple questions...........The question didn't explicitly state anything about the $M$. All I've done up to this point is prove the usual properties of $\limsup$ and $\liminf$ that most analysis textbooks have you start out with. – dc3rd Jun 19 at 23:51
• What is the definition of $\limsup$ that you are working with? – José Carlos Santos Jun 19 at 23:54